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Let $K$ be algebraically closed field. Can we show there are no non-constant morphisms from $\mathbb{P}^2 \to \mathbb{P}^1$? What is the most elementary way to show this (assuming its true)?

Sloan
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Eins Null
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    this is always the case when you have $n > m$ for a morphism of the form $\mathbb{P}^{n} \rightarrow \mathbb{P}^{m}$ –  Sep 19 '16 at 15:54
  • Whats an argument to show this? – Eins Null Sep 19 '16 at 15:55
  • if you are familiar of line bundles then you might get a more concrete answer –  Sep 19 '16 at 15:55
  • yeah you're right it's the same. the answer given by guy called Jesko is what I am talking to. –  Sep 19 '16 at 15:57
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    I'm thinking the problem is that somebody asking this question wouldn't know the correspondence between line bundles and sections and maps to projective space, and it'd help to have a more direct proof... – DCT Sep 19 '16 at 17:12

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As stated in the comments, using the correspondence between line bundles and maps to projective space, you can show that there is no map $\mathbb{P}^n\rightarrow \mathbb{P}^m$ for $n>m$ other than the constant map.

Since I'm curious, let me try to do this directly, and hopefully other people will find better ways.

Let $\phi:\mathbb{P}^2\dashrightarrow \mathbb{P}^1$ be a rational map. This is equivalent to a map of function fields going the other way. If we identify $K(\mathbb{P}^1)$ with $k(t)$ and $K(\mathbb{P}^2)$ with rational functions that are quotients of homogenous polynomials in $X,Y,Z$ of the same degree, then $\phi$ is induced by $t\rightarrow \frac{f(X,Y,Z)}{g(X,Y,Z)}$ where $f$ and $g$ are homogenous polynomials in $X,Y,Z$ with $\deg(f)=\deg(g)$. Without loss of generality, we can assume $f$ and $g$ are relatively prime.

This means, we can define $\phi:\mathbb{P}^2\rightarrow \mathbb{P}^1$ by $[X:Y:Z]\rightarrow [f(X,Y,Z):g(X,Y,Z)]$ on the complement of the set $S$ where $S=V(f,g)$.

I claim that this map cannot be extended over the set $S\subset\mathbb{P}^2$. First, since $f$ and $g$ have no common factors, $\dim(S)=0$. By Bezout's theorem, $S$ is nonempty. For $q\in \mathbb{P}^1$, let $\phi^{-1}(q)$ be the inverse image of $q$ under the map $\mathbb{P}^2\backslash S\rightarrow \mathbb{P}^1$.

Key Claim: Every $S$ is in the closure of $\pi^{-1}(q)$ for every $q\in \mathbb{P}^1$.

Let $q=[a:b]\in\mathbb{P}^1$. Then, the zero locus $V(bf(X,Y,Z)-ag(X,Y,Z))\subset\mathbb{P}^2$ contains both $\pi^{-1}(q)$ and $S$. Morever, $V(bf(X,Y,Z)-ag(X,Y,Z))\backslash S=\pi^{-1}(q)$. This means $\pi^{-1}(q)$ is dense in $V(bf(X,Y,Z)-ag(X,Y,Z))$. Since $V(bf(X,Y,Z)-ag(X,Y,Z))$ is pure of dimension 1, this means $S$ is in the closure of $\pi^{-1}(q)$.

Finally, having proved this key claim, we now know that $\phi$ cannot be extended over any $p\in S$, as this would contradict continuity.

DCT
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  • Dear DCT, I'm kinda confused as to how the contradiction is derived. So if $\phi$ could be extended over any $p \in S$, then we would have a continuous map $\phi$ from $\mathbb{P}^2 \ S \cup { p }$. And by continuity of $\phi$, for any $q \in \mathbb{P}^1$ , we must have $\phi^{-1}(q)$ is closed in $\mathbb{P}^2 \ S \cup { p }$? How to proceed from here? – eatfood Sep 09 '20 at 16:20
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    @eatfood I haven't checked the Key Claim again carefully, but the idea is that $\phi$ cannot be extended over any $p\in S$. The local picture to keep in mind is $\mathbb{C}^2\backslash {0}\rightarrow \mathbb{P}^1$ where $(x,y)\mapsto [x:y]$. Then, $(0,0)$ doesn't know where to go on $\mathbb{P}^1$. – DCT Sep 10 '20 at 15:28